if $\pi_{n}=(x-x_{1})...(x-x_{n})$ with $x_{i} \neq x_{j}$ for $i\neq j$, and $x_{i} \in [a,b]$ $a,b$ are reals how Can I prove that $\{\pi_{n}\}$ is a sequence of orthogonal polynomials by the following inner product:
$(\pi_{n},\pi_{m})=\displaystyle\int_{a}^{b} \pi_{n} \pi_{m} dx$.
Okay, I just have to prove that $(\pi_{n},\pi_{m})=0$ if $n\neq m$, but i have no idea on how solve this integral.
The statement is not true: take $$\pi_1(x)=x-\frac{1}{3}$$ and $$\pi_2(x)=\left(x-\frac{1}{3}\right)\left(x-\frac{1}{2}\right)$$ for $0\leq x\leq 1$, then $$(\pi_1,\pi_2)=\int_0^1\left(x-\frac{1}{3}\right)^2\left(x-\frac{1}{2}\right)dx=\frac{1}{36}\neq0,$$ as can be confirmed here
Therefore, these polynomials are not orthogonal with respect to the given inner product